Feedback Control Systems
Lecture 6
Dr.-Ing. Erwin Sitompul
President University
http://zitompul.wordpress.com
2 0 1 3
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Erwin Sitompul
FCS 6/1
Feedback Control Systems
Chapter 5
The Root-Locus Design Method
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FCS 6/2
Chapter 5
The Root-Locus Design Method
Root Locus: Illustrative Example
 Examine the following closed-loop system, with unity
negative feedback.
 The closed-loop transfer
function is given as:
 The roots of the characteristic
equation are:
Y ( s)
K
 2
R( s ) s  s  K
s1,2
1  1  4 K

2
• The characteristic equation
• The denominator of the
closed-loop transfer function
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Chapter 5
The Root-Locus Design Method
Root Locus: Illustrative Example
s1,2
 1
1  4K
1


,
0

K



2
4
 2
 1  j 4 K  1 , K  1
 2
2
4
Imaginary Axis
2
K=∞
K=1/4
1
0
: the poles of openloop transfer function
K=0
K=0
-1
-2
-2
K=∞
-1
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Real Axis
0
1
Erwin Sitompul
FCS 6/4
Chapter 5
The Root-Locus Design Method
Root Locus: Illustrative Example
 Where are the location of the closed-loop roots when K=1?
n2  1  n  1
2n  1    0.5
Y ( s)
K
1
 2
 2
R( s ) s  s  K s  s  1
  n  0.5
d  n 1   2  0.866
1.5
K=1
Imaginary Axis
1
0.866
(  jd )  (0.5  j 0.866)
0.5
0
-0.5
-1
-1.5
-2
K=1
-1
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–0.5
Real Axis
–0.866
0
1
There is a relation
between gain K and the
position of closed-loop
poles, which also affects
the dynamic properties
of the system (ζ and ωd).
Erwin Sitompul
FCS 6/5
Chapter 5
The Root-Locus Design Method
Root Locus of a Basic Feedback System
 The closed-loop transfer function of the basic feedback
system above is:
Y ( s)
D( s)G( s)
 T ( s) 
R( s )
1  D( s)G( s) H ( s)
 The characteristic equation, whose roots are the poles of this
transfer function, is:
1  D( s)G ( s) H ( s)  0
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Chapter 5
The Root-Locus Design Method
Root Locus of a Basic Feedback System
 To put the characteristic equation in a form suitable for study
of the roots as a parameter changes, it is rewritten as:
1  KL( s )  0
where
KL( s)  D( s)G ( s) H ( s)
b( s )
L( s ) 
a(s)
 K is the gain of controller-plant-sensor combination.
 K is selected as the parameter of interest.
 W. R. Evans (in 1948, at the age of 28) suggested to plot the
locus (location) of all possible roots of the characteristic
equation as K varies from zero to infinity  root locus plot.
 The resulting plot is to be used as an aid in selecting the best
value of K.
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FCS 6/7
Chapter 5
The Root-Locus Design Method
Root Locus of a Basic Feedback System
 The root locus problem shall now be expressed in several
equivalent but useful ways.
1  KL( s )  0
b( s )
1 K
0
a( s)
a( s)  Kb( s )  0
1
L( s )  
K
 The equations above are sometimes referred to as “the root
locus form of a characteristic equation.”
 The root locus is the set of values of s for which the above
equations hold for some positive real value of K.
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Chapter 5
The Root-Locus Design Method
Root Locus of a Basic Feedback System
 Explicit solutions are difficult to obtain for higher-order
system  General rules for the construction of a root locus
were developed by Evans.
 With the availability of MATLAB, plotting a root locus
becomes very easy. It can be done simply by using the
command “rlocus(num,den)”.
 However, in control design we are also interested in how to
modify the dynamic response so that a system can meet the
specifications for good control performance.
 For this purpose, it is very useful to be able to roughly sketch
a root locus which will be used to examine a system and to
evaluate the consequences of possible compensation
alternatives.
 Also, it is important to be able to quickly evaluate the
correctness of a MATLAB-generated locus to verify that what
is plotted is in fact what was meant to be plotted.
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Chapter 5
The Root-Locus Design Method
Guidelines for Sketching a Human Face
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Chapter 5
The Root-Locus Design Method
Guideines for Sketching a Root Locus
 Deriving using the root locus form of characteristic equation,
1  KL( s )  0
1
L( s )  
K
b( s )
1

a( s)
K
 Taking the polynomial a(s) and b(s) to be monic, i.e., the
coefficient of the highest power of s equals 1, they can be
factorized as:
( s  z1 )( s  z2 ) ( s  zm )
1

( s  p1 )( s  p2 ) ( s  pn )
K
 If any s = s0 fulfills the equation above, then s0 is said to be
on the root locus.
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Chapter 5
The Root-Locus Design Method
Guideines for Sketching a Root Locus
 The magnitude condition implies:
( s  z1 ) ( s  z2 )
( s  zm )
( s  p1 ) ( s  p2 )
( s  pn )
s  s0
1
1
 
K K
Magnitude
Condition
 The phase condition implies:
( s  z1 )(s  z2 ) (s  zm )
1

    180
( s  p1 )( s  p2 ) ( s  pn ) s  s
K
Phase
Condition
0
 Defining ( s  zi )   i and ( s  pi )  i , the phase condition
can be rewritten as:
m
n
   
i 1
i
i 1
i
 180  360(l  1), l  0, 1, 2, 3,
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Chapter 5
The Root-Locus Design Method
Guideines for Sketching a Root Locus
“
The root locus is the set of values of s for which
1+KL(s) = 0 is satisfied as the real parameter K varies
from 0 to ∞. Typically, 1 + KL(s) = 0 is the
characteristic equation of the system, and in this case
the roots on the locus are the closed-loop poles of
that system.
”
“
The root locus of L(s) is the set of points in the s-plane
where the phase of L(s) is 180°. If the angle to a test
point from a zero is defined as ψi and the angle to a
test point from a pole as Φi, then the root locus of L(s)
is expressed as those points in the s-plane where, for
integer l, Σψi – ΣΦi =180° +360°(l–1).
”
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Chapter 5
The Root-Locus Design Method
Guideines for Sketching a Root Locus
 Consider the following
example.
s 1
L( s ) 
s( s  5) (s  2)2  4
: the poles of L(s)
: the zero of L(s)
s0  1  j 2
Test
point
 1  90
1  tan 1 (2 1)  116.6
2  0
3  tan 1 (4 1)  76.0
4  tan 1 (2 4)  26.6
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  
i
i
  1  1  2  3  4
 90 116.6  0  76.0  26.6
 129.2  180  360(l  1)
 s0 is not on the root locus
Erwin Sitompul
FCS 6/14
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 1:
The n branches of the locus start at the poles of L(s) and
m of these branches end on the zeros of L(s), while
n–m branches terminate at infinity along asymptotes.
Recollecting
( s  z1 ) ( s  z2 )
b( s ) 1
b( s )
1


 
a( s) K
( s  p1 ) ( s  p2 )
a( s)
K
( s  z1 ) ( s  z2 )
1
lim    lim
K 0 K
s  pi ( s  p ) ( s  p )
1
2
( s  zm )
( s  z1 ) ( s  z2 )
1
lim  0  lim
K  K
s  zi ( s  p ) ( s  p )
1
2
( s  zm )
( s  zm )
1

( s  pn ) K
( s  pn )
( s  pn )
 The root locus starts at K = 0 at the poles of L(s)
and ends at K = ∞ on the zeros of L(s)
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 2:
On the real axis, the loci (plural of locus) are to the left of
an odd number of poles and zeros.
5
4
3
2
1
: The root
locus
1
4
3
2
1
 Angles from real poles or zeros are 0° if the test point is to
the right and 180° if the test point is to the left of a given
pole or zero.
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FCS 6/16
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 The rule is now applied to obtain the root locus of:
1
L( s ) 
s ( s  4)2  16
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p1  0
p2,3  4  j 4
Erwin Sitompul
FCS 6/17
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 For any test point s0 on the real
axis, the angles φ1 and φ2 of two
complex conjugate poles cancel
each other, as would the angles
of two complex conjugate zeros
(see figure below).
1  tan 1 (2 4)  26.6
2  tan 1 (2 4)  26.6
1  2  0
3  0 The pair does not
give contribution to
the phase condition
1  2  3  0
 180  360(l  1)
s1
 s0 is not on the
root locus
Now, check the phase
condition of s1!
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 3:
For large s and K, n–m of the loci are asymptotic to
lines at angles Φl radiating out from the point
s = α on the real axis, where:
180  360(l  1)
l 
, l  1, 2,
nm
,3
Angles of Asymptotes
p  z


i
i
nm
Center of Asymptotes
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
1
 For L( s) 
, we obtain n  3, m  0
2
s ( s  4)  16
p1  0, p2,3  4  j 4
180  360(l  1)
l 
nm
180  360(l  1)

30
 60  120(l  1)
 60,180,300
60°
180°
–2.67
p  z


i
i
nm
(4  j 4)  (4  j 4)  0

30
8

 2.67
3
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300°
Erwin Sitompul
FCS 6/20
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 4:
The angle of departure of a branch of a locus from
a pole is given by:
l ,dep   i  i  180  360(l  1)
i l
and the angle of arrival of a branch of a locus to
a zero is given by:
 l ,arr  i   i  180  360(l  1)
i l
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 For the example, the root loci must depart with certain
angles from the complex conjugate poles at –4 ± j4, and go
to the zero at ∞ with the angles of asymptotes 60° and
300°.
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FCS 6/22
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 From the figure,
2   i  i  180  360(l  1)
i2
2  1  3  180  360(l  1)
 But
1  90
3  tan 1 (4 4)  135
 Thus
2  90  135  180
2  405  45
 By the complex conjugate symmetry of the roots, the angle
of departure of the locus from –4 – j4 will be +45°.
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 So, the root loci will start their journey from –4 ± j4 towards
∞ with the direction of ±45°.
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 5:
The locus crosses the jω axis (imaginary axis) at points
where:
 The Routh criterion shows a transition from roots in the left
half-plane to roots in the right half-plane.
 This transition means that the closed-loop system is
becoming unstable.
 This fact can be tested by Routh’s stability criterion,
with K as the parameter, where an incremental change
of K will cause the sign change of an element in the first
column of Routh’s array.
 The values of s = ± jω0 are the solution of the
characteristic equation in root locus form, 1 + KL(s) = 0.
 The points ± jω0 are the points of cross-over on the
imaginary axis.
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 For the example, the characteristic equation can be written
as:
1  KL( s )  0
1
1 K
0
2
s ( s  4)  16
s 3  8s 2  32s  K  0
s3 :
1
32
s2 :
8
K
8  32  K
s :
8
s0 :
K
1
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• The closed-loop system is stable for K > 0
and K < 256  for 0 < K < 256.
• For K > 256 there are 2 roots in the RHP
(two sign changes in the first column).
• For K = 256 the roots must be on the
imaginary axis.
Erwin Sitompul
FCS 6/26
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 The characteristic equation is now solved using K = 256.
s 3  8s 2  32s  256  0
s1  8
s2,3   j 5.66   j0
Points of Cross-over
 Another way to solve for ω0 is by simply replacing any s with
jω0 without finding the value of K first.
( j0 )3  8( j0 ) 2  32( j0 )  K  0
 j03  80 2  j 320  K  0
K  80 2  j (320  03 )  0
≡
0
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≡
0
320  03
K  80 2
 K  8  32
 0 2  32
 K  256
 0  5.66
Same results for K and ω0
Erwin Sitompul
FCS 6/27
Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 The points of cross-over are now inserted to the plot.
5.66
–5.66
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 The complete root locus plot can be shown as:
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
RULE 6:
The locus will have multiple roots at points
on the locus where:
da( s )
db( s )
b( s )
 a( s)
0
ds
ds
The branches will approach and depart
a point of q roots at angles separated by:
180  360(l  1)
q
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Chapter 5
The Root-Locus Design Method
Rules for Plotting a Root Locus
 A special case of point of multiple roots is the intersection
point of 2 roots that lies on the real axis.
 If the branches is leaving the real axis and entering the
complex plane, the point is called the break-away point.
 If the branches is leaving the complex plane and entering the
real axis, the point is called the break-in point.
Imag
axis
Imag
axis
90
90
Real
axis
Breakaway
point
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Real
axis
Break-in
point
Erwin Sitompul
FCS 6/31
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
Draw the root locus plot of the system shown below.
K
Y (s)
s ( s  1)( s  2)

K
R( s) 1 
s ( s  1)( s  2)
 1  KL( s )
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1
 L( s ) 
RULE 1
s( s  1)( s  2)
3 zeros at infinity
n  3, m  0
p1  0, p2  1, p3  2
Erwin Sitompul
FCS 6/32
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
3
p1  0, p2  1, p3  2
Imag
axis
2
1
RULE 2
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
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Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
180  360(l  1)
l 
nm
180  360(l  1)

30
 60  120(l  1)
 60,180,300
p  z


i
nm
0 1  2

30
 1
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i
Angles of Asymptotes
p1  0, p2  1, p3  2
Center of Asymptotes
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FCS 6/34
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
l  60,180,300
3
  1
2
Imag
axis
60°
1
180°
–4
–3
–2
–1
–1
RULE 4
Not applicable. The angles of
departure or the angles of
arrival must be calculated only
if there are any complex poles
or zeros.
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1
2
Real
axis
RULE 3
–2
300°
–3
FCS 6/35
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
1  KL( s )  0
1
1 K
0
s( s  1)( s  2)
s 3  3s 2  2s  K  0
Replacing s with jω0,
( j0 )3  3( j0 ) 2  2( j0 )  K  0
 j03  30 2  j 20  K  0
K  30 2  j (20  03 )  0
≡
0
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≡
0
20  03
K  30 2
 K  3 2
 0 2  2
K 6
 0  1.414
Points of Cross-over
Erwin Sitompul
FCS 6/36
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
3
0  1.414
Imag
axis
2
1.414
RULE 5
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–1.414
–2
–3
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Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
1  KL( s )  0
b( s )
1
1
L( s ) 

 3
a( s) s( s  1)( s  2) s  3s 2  2s
The root locus must have a break-away point on the real axis,
between –1 and 0. It can be found by solving:
da( s )
db( s )
b( s )
 a( s)
0
ds
ds
1 (3s 2  6s  2)  ( s3  3s 2  2s)  0  0
3s 2  6s  2  0
s1  1.577, s2  0.423
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Erwin Sitompul
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Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
s1  1.577, s2  0.423
Not on the
root locus
3
On the root locus
The break-away
point
2
–0.423
–4
–3
–2
Imag
axis
–1
1.414
RULE 6
1
0
1
2
Real
axis
–1
–1.414
–2
–3
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Erwin Sitompul
FCS 6/39
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
After examining RULE 1 up to
RULE 6, now there is enough
information to draw the root
locus plot.
3
Imag
axis
2
–0.423
1.414
1
90
–4
–3
–2
–1
0
1
2
Real
axis
–1
–1.414
–2
–3
President University
Erwin Sitompul
FCS 6/40
Chapter 5
The Root-Locus Design Method
Example 1: Plotting a Root Locus
The final sketch, with direction of
root movements as K increases
from 0 to ∞ can be shown as:
3
Imag
axis
Final Result
2
–0.423
–4
–3
–2
–1
1.414
1
0
1
2
Real
axis
–1
–1.414
–2
Determine the locus of
all roots when K = 6!
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Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
a) Draw the root locus plot of the system.
b) Define the value of K where the system is stable.
c) Find the value of K so that the system has a root at s = –2.
President University
Erwin Sitompul
FCS 6/42
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
1
K
RULE 1
s2
Y (s)
( s  4)
 L( s ) 

s4
R( s ) 1  K 1 ( s  2)
There is one branch,
starts from the pole and
( s  4)
approaches the zero
n  1, m  1
p1  4, z1  2
 1  KL( s )
President University
Erwin Sitompul
FCS 6/43
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
3
p1  4, z1  2
Imag
axis
2
1
RULE 2
–4
–3
–2
–1
RULE 3
Not applicable, since n = m.
1
2
–1
–2
RULE 4
Not applicable. The angles
of departure or the angles
of arrival must be calculated
only if there are any
complex poles or zeros.
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0
Real
axis
Erwin Sitompul
–3
FCS 6/44
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
1  KL( s )  0
s2
1 K
0
s4
s  4  K ( s  2)  0
Replacing s with jω0,
( j0 )  4  K ( j0  2)  0
4  2 K  j0 (1  K )  0
≡0
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4  2K
K 2
≡0
0 (1  K )
 0  0
Points of Cross-over
Erwin Sitompul
FCS 6/45
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
RULE 5
3
The point of cross-over, as
can readily be guessed, is
at s = 0.
2
K=2
–4
Imag
axis
–3
–2
–1
1
0
1
2
Real
axis
–1
RULE 6
Not applicable. There is no
break-in or break-away point.
President University
Erwin Sitompul
–2
–3
FCS 6/46
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
a) Draw the root locus plot of the system.
3
The final sketch, with direction of
root movements as K increases
from 0 to ∞ can be shown as:
Imag
axis
Final Result
2
1
–4
–3
–2
–1
0
1
2
Real
axis
–1
–2
–3
President University
Erwin Sitompul
FCS 6/47
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
b) Define the value of K where the
system is stable.
3
Imag
axis
2
K=0
–4
K=2
–3
–2
–1
K=∞
1
0
1
2
Real
axis
–1
System is stable when the root of
the characteristic equation is on
the LHP, that is when 0 ≤ K < 2.
–2
–3
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Erwin Sitompul
FCS 6/48
Chapter 5
The Root-Locus Design Method
Example 2: Plotting a Root Locus
c) Find the value of K so that the
system has a root at s = –2.
3
2
K = 0.5
K=0
–4
K=2
–3
Imag
axis
–2
–1
K=∞
1
0
1
2
Real
axis
–1
Inserting the value of s = –2 in
the characteristic equation,
s  4  K ( s  2) s 2  0
2  4  K (2  2)  0
K (2  2)  2 4  0.5
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Chapter 5
The Root-Locus Design Method
Homework 6
 No.1, FPE (6th Ed.), 5.2.
Hint: Easier way is to assign reasonable values for the zeros
and poles in each figure. Later, use MATLAB to draw the root
locus.
 No.2, FPE (6th Ed.), 5.7.(b)
Hint: After completing the hand sketch, verify your result
using MATLAB. Try to play around with Data Cursor.
 No.3
Sketch the root locus diagram of the following closed-loop
system as accurate as possible.
 Due: 06.11.2013.
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